Are rationals adjoin cube root of 3 a field?

[Ragnarok7]

Is $$\mathbb{Q}(\sqrt[3]{3})=\{a+b\sqrt[3]{3}+c\sqrt[3]{9}\mid a,b,c\in\mathbb{Q}\}$$ a ring? If it is a ring, is it a field?

I have shown that it is a ring; however, I am not sure that it is a field, since in my calculations it does not seem to be closed under inverses. But I read somewhere that $$\mathbb{Q}(\sqrt[3]{2})$$ is a field, so could someone confirm or deny this for me? Thanks!

 

[Euge]

Is $$\mathbb{Q}(\sqrt[3]{3})=\{a+b\sqrt[3]{3}+c\sqrt[3]{9}\mid a,b,c\in\mathbb{Q}\}$$ a ring? If it is a ring, is it a field?

I have shown that it is a ring; however, I am not sure that it is a field, since in my calculations it does not seem to be closed under inverses. But I read somewhere that $$\mathbb{Q}(\sqrt[3]{2})$$ is a field, so could someone confirm or deny this for me? Thanks!

Hi Ragnarok,

Indeed $\Bbb Q(\sqrt[3]{3})$ is a field. Let $\alpha = \sqrt[3]{3}$. Then $\alpha$ is a root of the polynomial $x^3 - 3$, so we can identify $\Bbb Q(\sqrt[3]{3})$ with polynomials $p(x)\in Q[x]$ subject to the constraint $x^3 = 3$. This is naturally identified with the factor ring $\Bbb Q[x]/(x^3 - 3)$. Rigorously speaking, since the substitution map $\phi_{\sqrt[3]{3}} : \Bbb Q[x] \to \Bbb Q(\sqrt[3]{3})$ sending $p(x)$ to $p(\sqrt[3]{3})$ is a ring homomorphism with kernel $(x^3 - 3)$, it induces an isomorphism of $\Bbb Q[x]/(x^3 - 3)$ onto $\Bbb Q(\sqrt[3]{3})$. Since $x^3 - 3$ is irreducible over $\Bbb Q$, the ideal $(x^3 - 3)$ is maximal in $\Bbb Q[x]$ and hence $\Bbb Q[x]/(x^3 - 3)$ is a field. Therefore $\Bbb Q(\sqrt[3]{3})$ is a field.

 

[mathbalarka]

(more or less Euge's answer, a bit simplified and nonrigorified)

Let us think of it in a more general context : $F$ be a field and $\alpha$ be some $F$-algebraic (root of a polynomial over $F[x]$) of degree $n$. Define the ring $F[\alpha]$ to be the ring with arbitrary elements of the form $a_0 + a_1 \alpha^1 + a_2 \alpha^2 + \cdots + a_{n - 1} \alpha^{n-1}$ for $a_i \in F$.

It is, as you see, most certainly a ring. We claim that the ring $F[\alpha]$ is already closed under inversion and thus is a field. Let's look at some examples first : Take the field $\Bbb Q$. $\sqrt{2}$ is an algebraic over the field (root of $X^2 - 2$), thus adjoin it to the base field to get the ring $\Bbb Q[\sqrt{2}]$. The arbitrary elements here are of the form $a + b\sqrt{2}$, as $\sqrt{2}$ is a quadratic. If this ring is to be a field, each element of this form must have an inverse. However, note that $(a + b\sqrt{2})(a - b\sqrt{2}) = a^2 - 2b^2 \in \Bbb Q$. Thus, the inverse of $a + b\sqrt{2}$ is $a' + b'\sqrt{2}$, where $a' = a(a^2-2b^2)^{-1}$ and $b' = -b(a^2-2b^2)^{-1}$. Hence $\Bbb Q[\sqrt{2}]$ is a field.

A powerful way to think of field extensions is to compare with polynomial rings. Take $F[\alpha]$ and let's compare it with $F[X]$ for some $F$-transcendental $X$. There is a canonical map $\varphi : F[X] \to F[\alpha]$ by sending $X$ to $\alpha$. It's clearly an injective homomorphism, however it's NOT surjective. $0$ in $F[X]$ goes to $0$ in $F[\sqrt{2}]$, but the minimal polynomial $f(X)$ of $\alpha$ in $F[X]$ also goes to $0 = f(\alpha)$ in $F[\alpha]$. In fact, all $F[X]$-multiples of $f(X)$ goes to the identity in $F[\alpha]$, so the measure of nonsurjectivity (the fancy name being "kernel") is the ideal $\langle f(X) \rangle = \{f(X) \cdot g(X) \, : \, g(X) \in F[X] \}$. From the first isomorphism theorem $F[\alpha] \cong F[X]/\langle f(X) \rangle$.

Note that $f(X)$ is an irreducible polynomialso if one picks up some arbitrary polynomial $k(X)$ not included in the fibres over $0$ w.r.t. $\varphi$, then $(f(X), k(X)) = 1$ and by Bezout's lemma, there exists polynomials $a(X)$ and $b(X)$ such that $a(X)k(X) + b(X)f(X) = 1$. The identity must also hold modulo $f(x)$, so $a(x)k(x) = 1$ over $F[X]/\langle f(X) \rangle$, thus $a(X)$ is the inverse of $k(X)$, implying $F[X]/\langle f(X) \rangle$ is a field. Hence $F[\alpha]$ is also a field. $\blacksquare$

 

[Ragnarok7]

Thank you mathbalarka, for the explanation! I was surprised at how much I understood of it.